Answer:
Part A: The speed the car should be travelling when leaping the river is approximately 37.948 m/s
Part B: The speed of the car just before it lands is approximately 41.92345 m/s
Step-by-step explanation:
The parameters of the car attempting leaping the river are;
The height of the car over the river = 18 m
The height of the opposite side of the bridge above the river = 1.8 m
The width of the river, x = 69.0 m
Part A
The time, 't' it would take the car to fall from 18 m above the river to 1.8 m above the river is given as follows;
t = √(2·h/g)
Where;
h = The height of the fall = 18 m - 1.8 m = 16.2 m
g = The acceleration due to gravity = 9.8 m/s²
∴ t = √(2×16.2 m/(9.8 m/s²)) = (9/7)·√2 s
The horizontal speed, 'vₓ', with which the car should be travelling at the time it leaves the road in order just to clear the river and land safely on the opposite side is given as follows;
vₓ = x/t = 69.0 m/((9/7)·√2 s) = (161/6)·√2 m/s ≈ 37.948 m/s
The horizontal speed the car should be travelling when leaping the river, vₓ ≈ 37.948 m/s
Part B;
The vertical velocity of the car is given as follows;
² =
² + 2·g·h
∴
² =2·g·h = 2 × 9.8 m/s² × 16.200 m = 317.52m²/s²
= √(317.52 m²/s²) = (63/5)·√2 ≈ 17.819 m/s
The magnitude of the speed of the car, 'v', just before it lands is given using Pythagoras' theorem for resultant vectors as follows;
v = √(
² + vₓ²) = √(317.52 m²/s² +((161/6)·√2 m/s)²) ≈ 41.92345 m/s
The speed of the car just before it lands, v ≈ 41.92345 m/s.